Problem: Let $R$ be the region enclosed by the line $y=4$, the $y$ -axis, and the curve $y=\dfrac12 x^3$. $y$ $x$ ${y=\dfrac12 x^3}$ ${y=4}$ $y=-1}$ $ 0$ $(2,4)$ $ R$ A solid is generated by rotating $R$ about the line $y=-1$. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Answer: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=\dfrac12 x^3}$ ${y=4}$ $y=-1}$ $ 0$ $(2,4)$ Let the thickness of each slice be $dx$, let the radius of the washer, as a function of $x$, be $r_1(x)$, and let the radius of the hole, as a function of $x$, be $r_2(x)$. Then, the volume of each slice is $\pi[(r_1(x))^2-(r_2(x))^2]\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(x))^2-(r_2(x))^2]\,dx$ This is called the washer method. What we now need is to figure out the expressions of $r_1(x)$ and $r_2(x)$, and the interval of integration. $r_1(x)$ is equal to the distance between line $y=4$ and the line $y=-1$. So, ${r_1(x)=5}$. $r_2(x)$ is equal to the distance between the curve $y=\dfrac12 x^3$ and the line $y=-1$. So, ${r_2(x)=\dfrac12 x^3+1}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(x)})^2-({r_2(x)})^2] \\\\ &= \pi\left[ ({5})^2-\left( {\dfrac12 x^3+1} \right)^2 \right] \\\\ &=\pi\left[ 25-\left( \dfrac14 x^6+x^3+1 \right) \right] \\\\ &=\pi\left( -\dfrac14 x^6-x^3+24 \right) \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=2$. So the interval of integration is $[0,2]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^2 \left[ \pi\left( -\dfrac14 x^6-x^3+24 \right) \right]dx \\\\ &=\pi \int_0^2 \left( -\dfrac14 x^6-x^3+24 \right)\, dx \end{aligned}$ Let's evaluate the integral. $\pi \int_0^2 \left( -\dfrac14 x^6-x^3+24 \right)\, dx=\dfrac{276\pi}{7}$ In conclusion, the volume of the solid is $\dfrac{276\pi}{7}$.